It only remains to verify the identities. Consider 8.7 first. Fixing y and considering both
sides as a function of x, it follows from the above that both sides of the identity satisfy the
initial value problem

y′′ + y = 0,y (0) = sin(y),y′(0) = cos(y)

Therefore, the difference satisfies the initial value problem of Lemma 8.3.3. Therefore, by this
lemma, the difference equals 0. The next identity is handled similarly. This proves the
theorem.

Using 8.6, it follows sin is also periodic of period 2π. This proves the theorem.

Note that 2π is the smallest period for these functions. This can be seen by observing that
the above theorem and proof imply that cos is positive on

( )
(0, π-), 3π,2π
2 2

and negative on

(π2, 32π)

and that similar observations on sin are valid. Also, by
considering where these functions are equal to 0, 1, and -1 along with where they are
positive and negative, it follows that whenever a2 + b2 = 1, there exists a unique
t ∈ [0,2π) such that cos

(t)

= a and sin

(t)

= b. For example, if a and b are both
positive, then since cos is continuous and strictly decreases from 1 to 0 on

[0, π2]

, it
follows there exists a unique t ∈

(0,π ∕2)

such that cos

(t)

= a. Since b > 0 and sin is
positive on

(0,π∕2)

, it follows sin

(t)

= b. No other value of t in [0,2π) will work
since only on

(0,π∕2)

are both cos and sin positive. If a > 0 and b < 0 similar
reasoning will show there exists a unique t ∈ [0,2π) with cos

(t)

= a and sin

(t)

= b
and in this case, t ∈

(3π∕2,2π)

. Other cases are similar and are left to the reader.
Thus, every point on the unit circle is of the form

(cost,sin t)

for a unique t ∈ [0,
2π).

This shows the unit circle is a smooth curve, however this notion will not be considered
here.

Corollary 8.3.8For all x ∈ ℝ

sin(x+ 2π) = sin(x), cos(x+ 2π) = cos(x)

Proof: Let y

(x)

≡ sin

(x + 2π)

− sin

(x)

. Then from what has been shown above,
y′

(0)

= y

(0)

= 0. It is also clear from the above that y′′ + y = 0. Therefore, from Lemma
8.3.3y = 0. Differentiating the identity just obtained yields the second identity.
■

Are these the same as the circular functions you studied very sloppily in calculus and
trigonometry? They are.

If sin

(x)

defined above and sin

(x )

studied in a beginning calculus class both satisfy the
initial value problem

y′′ +y = 0,y(0) = 0,y′(0) = 1

then they must be the same. However, if you remember anything from calculus you will
realize sin

(x)

used there does satisfy the above initial value problem. If you don’t remember
anything from calculus, then it does not matter about harmonizing the functions. Just use the
definition given above in terms of a power series. Similar considerations apply to
cos.

Using the techniques of differentiation, you can find the derivatives of all these.

Now it is time to consider the exponential function exp

(x)

defined above. To do this, it is
convenient to have the following uniqueness theorem.

Lemma 8.3.9Suppose

y′ − y = 0,y(0) = 0

Then y = 0. Also for all x ∈ ℝ,exp(−x)(exp(x)) = 1

Proof: The function exp has been defined above in terms of a power series. From this
power series and Theorem 8.2.1 it follows that exp solves the above initial value problem.
Multiply both sides of the differential equation by exp

(− x)

. Then using the chain rule and
product rule,

-d-(exp(− x)y (x)) = 0
dx

and so exp

(− x)

y

(x)

= C, a constant. The constant can only be 0 because of the initial
condition. Therefore,

exp (− x)y(x) = 0

for all x.

Now I claim exp

(− x)

and exp

(x)

are never equal to 0. This is because by the chain rule,
abusing notation slightly,

′
(exp (− x)exp(x)) = − exp (− x)exp(x)+ exp(− x)exp(x) = 0

and so

exp(− x)exp(x) = C

a constant. However, this constant can only be 1 because this is what it is when
x = 0, a fact which follows right away from the definition in terms of power series.
■

Theorem 8.3.10The function exp satisfies the following properties.

exp

(x)

> 0 for all x ∈ ℝ, limx→∞exp

(x)

= ∞,limx→−∞exp

(x)

= 0.

exp is the unique solution to the initial value problem

y′− y = 0,y(0) = 1 (8.9)

(8.9)

For all x,y ∈ F

exp(x+ y) = exp (x )exp (y) (8.10)

(8.10)

exp is one to one mapping ℝ onto

(0,∞ )

.

Proof: To begin with consider 8.10. Fixing y it follows from the chain rule and the
definition using power series that

x → exp(x + y)− exp (x)exp(y)

satisfies the initial value problem of Lemma 8.3.9 and so it is 0. This shows 8.10.

8.9 has already been noted. It comes directly from the definition and was proved in
Lemma 8.3.9. The claim that exp